2023 CGSS Prelim 4049 P2 Solutions

CEDAR GIRLS’ SECONDARY SCHOOL Preliminary Examination 2023 Secondary Four CANDIDATE NAME SOLUTIONS CLASS CLASS INDEX NUMBER CENTRE/ INDEX NO / ADDITIONAL MATHEMATICS 4049/02 Paper 2 11 September 2023 2 hours 15 minutes Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your centre number, index number and name in the spaces at the top of this page. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. The use of an approved scientific calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 90. For Examiner’s Use 90 This document consists of 18 printed pages [Turn over

Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation 0 2 =++ cbxax , a acbbx 2 4 2 −−= Binomial expansion nrrnnnnn bbar nbanbanaba ++     ++     +     +=+ −−− ......21)( 221 , where n is a positive integer and ! )1(...)1( )!(! ! r rnnn rnr n r n +−−=−=      2. TRIGONOMETRY Identities 1cossin 22 =+ AA AA 22 tan1sec += 22cosec 1 cotAA=+ BABABA sincoscossin)sin( = BABABA sinsincoscos)cos( = BA BABA tantan1 tantan)tan(  = AAA cossin22sin = AAAAA 2222 sin211cos2sincos2cos −=−=−= A AA 2 tan1 tan22tan −= Formulae for ABC C c B b A a sinsinsin == Abccba cos2 222 −+= 1 sin2 bc A=

3 Cedar Girls’ Secondary School 4049/02/S4/Prelims/2023 [Turn over Answer all the questions. 1 The mass, x grams, of a volatile matter from a space mission remaining t days after being exposed to Earth’s atmosphere is given by 0.51.3 7 txe −=+ . (a) Find the initial mass of the matter. [1] (b) Explain why the mass of the substance can never be lower than 1.3 grams. [2] (c) Find the least number of days it takes for the matter to be reduced to half of its initial mass. [3] t=0, 1.3 7 8.3x= + = grams For all real values of t ( 0t ), 0.5 0te−  0.570 te−  0.57 1.3 1.3te− + Hence the lowest value will be 1.3 grams Half of initial mass 8.3 2 4.15=  = grams 0.54.15 1.3 7 te−=+ 0.557 140 te−= 57ln 0.5140 t=− 57ln 0.5140t = − 1.80t = days (3 s.f)

4 Cedar Girls’ Secondary School 4049/02/S4/Prelims/2023 2 (a) Prove that 211 2cotsec 1 sec 1 xxx −=−+ . [3] (b) Hence solve 11 5cossec 1 sec 1 ecxxx −=−+ , for 0 360x   . [5] 11 sec 1 sec 1LHS xx=− −+ 2 sec 1 (sec 1) sec 1 xx x + − −= − 2 2 tan x= 22cot x= 22cot 5cosx ecx= 22( sec 1) 5cosco x ecx−= 22 sec 5cos 2 0co x ecx− − = 5 25 4(2)( 2)cos 4ecx  − −= 4sin 5 41 x= + or 4sin 5 41 x= − Reference angle = 20.545 20.5 ,159.5x=  

5 Cedar Girls’ Secondary School 4049/02/S4/Prelims/2023 [Turn over 3 The polynomial 32f ( ) 5 3x ax bx x= + + − , where a and b are constants, is exactly divisible by 21x− and leaves a remainder of 39 when divided by 2x− . (a) Find the value of a and of b. [4] (b) Using these values of a and of b, determine the number of real roots of the equation f ( ) 0x = . Show all necessary working. [3] 32(0.5) (0.5) (0.5) 5(0.5) 3f a b = + + − 0 0.125 0.25 0.5ab= + − 0 2 4ab= + − 42ab=− - - - (1) 32(2) (2) (2) 5(2) 3f a b= + + − 39 8 4 7ab= + + 0 8 4 32ab= + − - - - (2) Sub (1) into (2), 0 8(4 2 ) 4 32bb= − + − 0 32 16 4 32bb= − + − 0b= 4a= 3f ( ) 4 5 3x x x= + − 2f ( ) (2 1)(2 3)x x x x= − + + 2(2 1)(2 3) 0x x x− + + = 2 4 1 4(2)(3) 23 0b ac− = − =−  Therefore there is only 1 real root, 1 2x=

6 Cedar Girls’ Secondary School 4049/02/S4/Prelims/2023 4 (a) If (4 3) 2 1y x x= − + , show that d 12 1 d 21 yx x x += + . [3] (b) Hence find the value of 12 3 d 21 x x x + + expressing your answer in the form 2 1( )x ax b++ where a and b are integers. [4] (4 3) 2 1y x x= − + 1 21(4) 2 1 (4 3)(2 1) (2)2 dy x x xdx − = + + − + (4 3)(4) 2 1 21 dy x xdx x −= + + + 4(2 1) (4 3) 21 dy x x dx x + + −= + 12 1 21 dy x dx x += + 12 3 12 1 2dd 2 1 2 1 2 1 xx x x C x x x ++ = + + + + + 2(4 3) 2 1 d 21 x x x C x = − + + + + 1 2(4 3) 2 1 2(2 1) dx x x x C − = − + + + +  1 22(2 1)(4 3) 2 1 1 22 xx x C += − + + +  (4 3) 2 1 2 2 1x x x C= − + + + + 2 1(4 1)x x C= + − +

7 Cedar Girls’ Secondary School 4049/02/S4/Prelims/2023 [Turn over 5 (a) The area of a quadrilateral is given as 25(tan15 ) cm2. Without using a calculator, express the area in the form ( 3ab+ ) cm2. [4] (b) Given that tan15 is a root to the equation 2 0x px q+ + = , where p and q are integers, find the value of p and q. [3] 25(tan15 ) = 25(tan(45 30 ))− = tan 45 tan 3025 1 tan 45 tan 30    −  + = 31 325 31 3  −   + = 33 325 33 3  −  + = 3325 33  − + = 3 3 3 325 3 3 3 3  −−  +− = 12 6 325 6  − = 50 25 3− tan15 2 3= − 2(2 3) (2 3) 0pq− + − + = 7 4 3 2 3 0 p p q− + − + = 7 2 4 3 3p q p+ − =− + 4p=− 72 pq+ =− 7 2( 4) q+ − =− 1q=

8 Cedar Girls’ Secondary School 4049/02/S4/Prelims/2023 6 The figure below consists of a rectangle ABCD and an isosceles triangle AED, where AB = y cm, BC = 2x cm and ED = 5 4 x cm. Given that the perimeter of ABCDE is 70 cm, (a) show that the area of figure is 21570 4A x x=− . [5] 52 2 2 4P x y x = + +  970 2 2 xy=+ 92 70 2 xy=− 140 9 4 xy −= Height = 225()4 xx − Height = 3 4 x 1322 24A xy x x  = +     140 9 32 44 xA x x x −   =+        229370 24A x x x= − + 21570 4A x x=− A C B D E 5 4 x cm y cm 2x cm

9 Cedar Girls’ Secondary School 4049/02/S4/Prelims/2023 [Turn over (b) Given that x can vary, find the value of x for which the area of the figure is at a maximum. [5] 140 9 4 xy −= d 15 70d2 A xx =− 150 70 2 x=− 19 3x= or 9.33 (3 s.f) 2 2 d 15 d2 A x =− By second derivative test, A is a maximum when 19 3x=

10 Cedar Girls’ Secondary School 4049/02/S4/Prelims/2023 7 (a) Solve the equation 0633 212 =+− ++ xx . [4] (b) Solve ( ) ( )2 2log 2 1 log 1x